Determining Convergence or Divergence of Improper Integrals: A Case Study

Improper integrals play a crucial role in mathematical analysis, particularly when dealing with functions that are not defined over a finite interval or have singularities. In this article, we will walk through a specific case of an improper integral to demonstrate how we can determine whether it converges or diverges. The integral in question is:

(int_{0}^{infty} frac{x ln{x}}{1 - x^2} dx)

Identifying the Discontinuity and Improper Integral Nature

First, let's examine the integrand: (frac{xln{x}}{1 - x^2}). Notice that the integrand has a removable discontinuity at (x 0), because:

[lim_{x to 0^ } x ln{x} 0]

Even though the integrand approaches infinity as (x to 0^ ), the discontinuity is removable, meaning the integral is only improper at infinity. Therefore, we can focus on the behavior of the integral as (x) approaches infinity.

Applying the Comparison Test

To determine whether the integral converges or diverges, we can use the Comparison Test. The Comparison Test states that if (0 leq f(x) leq g(x)) for all (x geq a), and if (int_a^infty g(x) dx) converges, then (int_a^infty f(x) dx) also converges. Conversely, if (int_a^infty f(x) dx) diverges, then (int_a^infty g(x) dx) diverges.

In our case, we observe that:

[ln{x} leq xtext{ for }x geq 0]

Let's manipulate the integrand to make the comparison more straightforward:

[frac{x ln{x}}{1 - x^2} leq frac{x(1 - x^2)}{1 - x^2} x]

Establishing Convergence Using Direct Comparison

Now, let's consider the integral:

[int_0^{infty} frac{x}{1 - x^2} dx]

For (0

[frac{x}{1 - x^2} leq frac{x}{1 - x}]

For (x geq 1), we have:

[frac{x}{1 - x^2} leq frac{x}{1 - x^2}]

Therefore, we need to analyze the integral:

[int_0^{infty} frac{x}{1 - x^2} dx]

This can be rewritten as:

[int_0^{infty} frac{x}{1 - x^2} dx -frac{1}{2} int_0^{infty} frac{1}{x^2 - 1} dx]

Let's break this into two parts:

[int_0^{infty} frac{x}{1 - x^2} dx -frac{1}{2} left( int_0^1 frac{1}{x^2 - 1} dx int_1^{infty} frac{1}{x^2 - 1} dx right)]

For the second integral, let's use the substitution (u x^2 - 1):

[du 2x dx Rightarrow dx frac{du}{2x}]

When (x 1), (u 0); and when (x to infty), (u to infty). Therefore:

[int_1^{infty} frac{1}{x^2 - 1} dx int_0^{infty} frac{1}{2u} du frac{1}{2} left[ ln|u| right]_0^{infty} infty]

Concluding the Convergence of the Original Integral

Given that the integral (int_1^{infty} frac{1}{x^2 - 1} dx) diverges, we conclude that:

[int_0^{infty} frac{x}{1 - x^2} dx]

diverges. However, from our earlier comparison:

[frac{x ln{x}}{1 - x^2} leq frac{x}{1 - x^2}]

Therefore, by the Comparison Test, we conclude that:

[int_0^{infty} frac{x ln{x}}{1 - x^2} dx]

converges.

Summary

In this article, we demonstrated how to determine the convergence or divergence of a specific improper integral. By identifying the nature of the integrand's discontinuity and applying the Comparison Test, we were able to show that the integral (int_0^{infty} frac{x ln{x}}{1 - x^2} dx) converges. This process exemplifies the importance of mathematical analysis in understanding the behavior of functions and integrals.

Keywords

improper integral, convergence, divergence, comparison test